3
$\begingroup$

To what I understand, the following is a valid way to introduce the angular momentum $\mathbf L$ in the Lagrangian system of a rigid body. We can consider the extended configuration space to be $M\times \mathbb{R}=SO(3)\times \mathbb R$. For a left invariant Riemannian metric $g$ in $SO(3)$ we can define the Lagrangian $\mathcal L:TM\to \mathbb R$ as the map $v\mapsto \frac 1 2 g(v,v)$. Suppose $v\in TM$, we can define the map $\Omega$ from TM into the Lie algebra $\mathfrak g=so(3)=T_eM$ of $SO(3)$ by $$v\mapsto \Omega_v=\big(L_{g^{-1}}\big)_*v\quad \text{for }v\in T_gM,$$ where $L_g$ is the left translation on $M$. Hence, the Lagrangian take the form $\mathcal L(v)=g(\Omega_v,\Omega_v)$. $\Omega_v$ is called the angular velocity. Furthermore, since the killing form $B:\mathfrak g\times \mathfrak g\to \mathbb R$ is a bilinear non-degenerate form on $so(3)$ (since it is a semisimple Lie algebra) defined by $B(\alpha,\beta)=-1/2 \; \text{Tr } (\alpha\beta)$. Then there exists a symmetric endomorphism $\mathbf L:\mathfrak g\to \mathfrak g$ with the property that $$\mathcal L(v)=B(\mathbf L\Omega_v,\Omega_v).$$ Such map (for fixed $v\in TM$ and thus fixed $\Omega_v$) is called the Angular momentum of the body. This is a nice and elegant way to define the Angular momentum, for the eigenvalues of $\mathbf L$ are the principal axes of inertia, and its corresponding eigenvalues, the principal moments of inertia. But what is still bothering me, is that the only physical intuition I get from such definition is the fact that a rigid body, can ultimately be characterized by translations and rotations (here I assume, translations are not being considered). But the definitions of the Lagrangian and of the angular momentum; seem to be too artificial to me. Is there a nice way to see this?

$\endgroup$
8
  • 1
    $\begingroup$ What is the $\mathbb{R}$ in the configuration space? Why isn’t it $\mathbb{R}^3$? $\endgroup$
    – G. Smith
    Commented Jul 15, 2020 at 1:20
  • 1
    $\begingroup$ @G.Smith The configuration space is just $M=SO(3)$; by $M\times \mathbb R$ I mean the extended configuration space, i.e. taking into account the `time' parameter. $\endgroup$ Commented Jul 15, 2020 at 1:47
  • 1
    $\begingroup$ If you start from the Lagrangian, it's possible to observe that there is rotational symmetry and obtain angular momentum using Noether's theorem. It seems like you're hoping to start from rotational symmetry, then derive the Lagrangian. In other words, a satisfying answer would be a proof of something like "$\mathcal{L}$ is the unique function with [properties] invariant under the action of $SO(3)$". Is that right? $\endgroup$
    – Daniel
    Commented Jul 15, 2020 at 3:33
  • $\begingroup$ Who considers time part of “configuration space”? If that's a thing, I haven't seen it. $\endgroup$
    – G. Smith
    Commented Jul 15, 2020 at 5:23
  • 1
    $\begingroup$ Check out this paper. It makes some argument (which I haven't read in full) that a bi-invariant metric should be used. Whether or not this argument is convincing, it seems likely that we could impose some reasonable-sounding conditions which force the metrc to be bi-invariant. It also seems clear that nontrivial dependence on the coordinates is forbidden by symmetry. To me, the remaining interesting question is why a bilinear form? Is there a time-scaling argument that rules out, say, $\mathcal{L} = g^2$? $\endgroup$
    – Daniel
    Commented Jul 15, 2020 at 15:45

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.